Wave Optics — CBSE Class 12 · 2017 Board Question with Solution

(a) In Young's double slit experiment, a monochromatic source of light S is kept equidistant from the slits

S_{1}

and

S_{2}

. Explain the formation of dark and bright fringes on the screen. (b) A beam of light consisting of two wavelengths,

650

nm and

520

nm, is used to obtain interference fringes in a Young's double-slit experiment. (i) Find the distance of the third bright fringe on the screen from the central maximum for wavelength

650

nm. (ii) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? Given: the separation between the slits is

4

mm and the distance between the screen and the plane of the slits is

1.2

Answer: (i) $x = 0.585$ mm.
(ii) $x = 0.78$ mm (where the 4th fringe of $650$ nm meets the 5th of $520$ nm).

Solution

(a) Since S is equidistant from

S_{1}

and

S_{2}

, the two slits act as coherent sources in phase. At the central point O the path difference is zero, giving a bright fringe. At a general point the path difference is

Δ = \frac{x d}{D}

; where

Δ = nλ

the waves arrive in phase (bright fringe), and where

Δ = (2 n - 1) \frac{λ}{2}

they arrive out of phase (dark fringe). This produces alternate bright and dark fringes.
(b)(i) Third bright fringe for

λ_{1} = 650

nm (

d = 4 \times 1 0^{- 3}

D = 1.2

m):

x = \frac{n λ _{1} D}{d} = \frac{3 \times 650 \times 1 0 ^{- 9} \times 1.2}{4 \times 1 0 ^{- 3}} = 5.85 \times 1 0^{- 4} m = 0.585 mm

(ii) Bright fringes coincide when

n_{1} λ_{1} = n_{2} λ_{2}

\frac{n _{1}}{n _{2}} = \frac{λ _{2}}{λ _{1}} = \frac{520}{650} = \frac{4}{5}

so the

4^{th}

fringe of

650

nm coincides with the

5^{th}

fringe of

520

nm:

x = \frac{4 \times 650 \times 1 0 ^{- 9} \times 1.2}{4 \times 1 0 ^{- 3}} = 7.8 \times 1 0^{- 4} m = 0.78 mm

x_{(i)} = 0.585 mm, x_{(ii)} = 0.78 mm

CBSE Class 12 Board · 2017 · Subjective

CBSE Class 12 · 2017 · Wave OpticsLong Answer

Solution

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